Challenging In-tegral

$\begin{aligned} I & = \int_0^1 \ln\left(\sqrt{1-x}+\sqrt{1+x}\right)\ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = x\ln\left(\sqrt{1-x}+\sqrt{1+x}\right)\ \bigg|_0^1 - \int_0^1 \frac {\frac x2\left(\frac {-1}{\sqrt{1-x}}+\frac 1{\sqrt{1+x}}\right)}{\sqrt{1-x}+\sqrt{1+x}}dx \\ & = \frac {\ln 2}2 - \int_0^1 \frac {x\left(\sqrt{1-x}-\sqrt{1+x}\right)}{2\sqrt{1-x^2}\left(\sqrt{1-x}+\sqrt{1+x}\right)} dx \\ & = \frac {\ln 2}2 - \int_0^1 \frac {x\left(\sqrt{1-x}-\sqrt{1+x}\right)^2}{2\sqrt{1-x^2}\left(1-x-1-x\right)} dx \\ & = \frac {\ln 2}2 + \frac 12 \int_0^1 \frac {1-\sqrt{1-x^2}}{\sqrt{1-x^2}} dx \\ & = \frac 12 \left(\ln 2 + {\color{#3D99F6}\int_0^1 \frac 1{\sqrt{1-x^2}} dx} - \int_0^1 dx \right) & \small \color{#3D99F6} \text{Let }x = \sin \theta \implies dx = \cos \theta \ d\theta \\ & = \frac 12 \left(\ln 2 + {\color{#3D99F6}\int_0^\frac \pi 2 d\theta} - 1 \right) \\ & = \frac 12 \left(\ln 2 + \frac \pi 2 - 1 \right) \end{aligned}$

Therefore, $a+b+c = 2+2-1=\boxed 3$ .